\newproblem{lay:1_9_39}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 1.9.39}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Let $T$ be a linear transformation whose standard matrix is given by \begin{center}$A=\begin{pmatrix} 4 & -7 & 3 & 7 & 5\\ 6 & -8 & 5 & 12 & -8\\-7 & 10 & -8 & -9 & 14 \\
	   3 & -5 & 4 & 2 & -6 \\ -5 & 6 & -6 & -7 & 3\end{pmatrix}$\end{center}
	Does $T$ map $\mathbb{R}^5$ onto $\mathbb{R}^5$?
}{
  % Solution
	The standard matrix is row-equivalent to
	\begin{center}
		$\begin{pmatrix} 1 & 0 & 0 & 5 & 0\\ 0 & 1 & 0 & 1 & 0\\0 & 0 & 1 & -2 & 0 \\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0\end{pmatrix}$
	\end{center}
	The transformation is not onto because there are only 4 pivot columns, i.e., only 4 linearly independent vectors and we need 5 to span $\mathbb{R}^5$.
}
\useproblem{lay:1_9_39}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
